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Chapter 2. Resistive circuits

2014. 9. 12.

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2Contents

1. Ohm’s law

2. Kirchhoff’s laws

3. Series and parallel resistor combinations

4. Y-Δ transformation

5. Circuits with dependent sources

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3

Entering resistive material, charges are decelerated, which decrease current flow.

Resistors : microscopic view

nucleus

electrons

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4Types of resistors

(1), (2), and (3) are high power resistors. (4) and (5) are high-wattage fixed resistors. (6) is a high precision resistor. (7)–(12) are fixed resistors with different power ratings.

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5

Rtit )()(

1. Ohm’s law

)0( R

RRitittp

22)()()( Power absorption :

resistance

RG 1

; conductance

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6Example 2.1

Determine the current and the power absorbed by the resistor.

][6212 mA

kI

kRV

kRI

WVIP

2/)12(/

)2()106(

][072.0)106)(12(

22

232

3

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7Glossary

(1) Node A node is simply a point of connection of two or more circuit elements.

node

Although one node can be spread out with perfect con-ductors, it is still only one node

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(3) branch

(2) loop A loop is simply any closed path through the circuit in which no nodeis encountered more than once

 a branch is a single or group of components such as resistors or a source which are connected between two nodes

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92. Kirchhoff’s law

(1) Kirchhoff ’s current law (KCL) : the algebraic sum of the currents entering(out-going) any node is zero→ the sum of incoming currents is equal to the sum of outgoing currents.

(2) Kirchhoff’s voltage law (KVL), the algebraic sum of the voltages around any loop is zero

0)()()( 54321 IIIII

54321 IIIII

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10Kirchhoff’s Current law0)(

nn tI

)(0 t

)(2 ti)(1 ti

)(3 ti3

303

2

202

1

101

321

-)(,-)(,-)(

0)()()()(

Rti

Rti

Rti

titititIn

n

Ri ba

ab

R

Current definition • The direction of a current can be chosen ar-bitrarily.

• The value of a current can be obtained from a voltage drop along the direction of current divided by a resistance met.

R2

)(1 t )(2 t

)(3 t

R1

R3

0---

3

30

2

20

1

10 RRR

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11Kirchhoff’s Voltage law

-(t)1 -(t)2

-(t)s

0)( n

n t

Sum of voltage drops along a closed loop should be equal to zero!

0)(-)()( 21 ttt s

R1 C1

Voltage convention

baab VVV

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12Example 2.6Find the unknown currents in the network.

020601 mmINode 1 :

Node 2 : 0614 III

Node 3 : 0406054 mmII

030205 mmINode 4 :

Node 5 : 030406 mmI

][801 mAI ][505 mAI ][106 mAI ][704 mAI

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13Example E2.6

Find the current ix in the circuits in the figure.

04410 mii xx

][4 mAix

01212010 mmii xx

][12 mAix

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Find Vad and Veb in the network in the figure.

Example E2.8

][266424 VVad

][102468 VVeb

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15Example 2.15

Given the following circuit, let us find I, Vbd and the power absorbed by the 30kΩ resistor. Finally, let us use voltage division to find Vbc .

0301220106 kIkIkI

][1.060

6 mAk

I

][101221220 VkIVbd

][3.0301001.030 6230 mWkkIP k

][2)6(4020

20 Vkk

kVbc

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16Series resistors

equivalent

NS RRRR 21

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17Parallel resistors

NP RRRR1111

21

equivalent

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18Example 2.19

Given the circuit, we wish to find the current in the 12-kΩ load resistor.

equivalent

][25.0)1(13

1)1(

121

41

121

mAmm

kk

kIL

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19Example 2.20We wish to determine the resistance at terminals A-B in the network in the figure.

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20Y-Δ transformation

equivalent

321

312 )(RRR

RRRRR ba

321

213 )(RRR

RRRRR cb

321

321 )(RRR

RRRRR ac

Y

RRRRRR

RRRRRR

RRRRRR

c

b

a

321

13

321

32

321

21

YR

RRRRRRR

RRRRRRRR

RRRRRRRR

a

accbba

c

accbba

b

accbba

3

2

1

ΔY

321 RRRR 3

RRY

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21Example 2.26Given the network in Fig. 2.36a, let us find the source current IS .

][661812

1812

kkkk

kkRa

][361812

618

kkkk

kkRb

][261812

612

kkkk

kkRc

][4612612

kkkkkRP

][2.146

12 mAkk

IS

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222.8 Circuits with dependent sourcesExample 2.27

052000312 111 kIIkI

][2523

121 mA

kkkI

][10510 VkIV

Let us determine the voltage Vo in the circuit in the figure.

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23Example 2.28

Given the circuit in the figure containing a current-controlled current source, let us find the voltage Vo.

kVIII

kVm ss

3,04

610 000

0560016

10 sss Vk

Vk

Vm

][8],[12 0 VVVVs

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24Example 2.30

An equivalent circuit for a FET common-source amplifier or BJT common-emitter amplifier can be modeled by the circuit shown in the figure. We wish to determine an expression for the gain of the amplifier, which is the ratio of the output voltage to the input voltage.

543 |||| RRRRL

02111 RiRii

ig RRR

21

2

Lgm Rg 0

21

20)(RR

RRgRg

gainG Lmi

gLm

i

GND can be arbitrarily set.

V0

V0

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Transistor

Transistor amplifier

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262.10 Application examplesExample 2.33 : The Wheatstone bridge circuit.

xx

x

RR

RR

RRR

RRR 2

3

1

231

3

1

32 R

RRRx